Friday, December 27, 2019

Behaviour of conjugacy by reduction modulo integers

Let AA and BMn(Z)B\in\mathrm{M}_n(\mathbf Z) be two square matrices with integer coefficients. Assume that they are conjugate by GLn(Z)\mathrm{GL}_n(\mathbf Z), namely, that there exists a matrix PGLn(Z)P\in\mathrm{GL}_n(\mathbf Z) such that B=P1APB=P^{-1}AP. Then we can reduce this relation modulo every integer d2d\geq 2 and obtain a similar relation between the images of AA and BB in Mn(Z/dZ)\mathrm M_n(\mathbf Z/d\mathbf Z).

Almost the same holds if AA and BB are only conjugate by GLn(Q)\mathrm{GL}_n(\mathbf Q), except for a few exceptions: we just need to take care to reduce the relation modulo integers dd that are coprime to the denominators of the coefficients of PP or of P1P^{-1}.

I was quite surprised at first to learn that the converse assertion is false. There are matrices AA and BB in M2(Z)\mathrm M_2(\mathbf Z) whose images modulo every integer d2d\geq 2 are conjugate, but which are not conjugate by a matrix in GL2(Z)\mathrm{GL}_2(\mathbf Z).

An example is given by Peter Stebe in his paper “Conjugacy separability of groups of integer matrices”, Proc. of the AMS, 32 (1), mars 1972, p. 1—7.
Namely, set
 A=( 188275 121177)=(1117+125111121116+1) A = \begin{pmatrix} 188 & 275 \\ 121 & 177 \end{pmatrix} = \begin{pmatrix} 11\cdot 17+1 & 25\cdot 11 \\ 11^2 & 11\cdot 16+1 \end{pmatrix}
and
\[ B = \begin{pmatrix} 188 & 11 \\ 3025 & 177 \end{pmatrix} =
\begin{pmatrix} 11\cdot 17+1 & 11 \\ 11^2\cdot 25 & 11\cdot 16+1 \end{pmatrix}. \]
These matrices AA and BB have integer coefficients, their determinant is 11, hence they belong to SL2(Z)\mathrm{SL}_2(\mathbf Z). They also have the same trace, hence the same characteristic polynomial, which is T2365T+1T^2-365T+1. The discriminant of this polynomial is 31123673\cdot 11^2\cdot 367. This implies that their complex eigenvalues are distinct, hence these matrices are diagonalizable over C\mathbf C, and are conjugate over C\mathbf C.

In the same way, we see that they remain conjugate modulo every prime number pp
that does not divide the discriminant. Modulo 33 and 1111, we check that both matrices become
conjugate to (1101)\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}, while they become conjugate to (1101)\begin{pmatrix}-1 & 1 \\ 0 & -1 \end{pmatrix} modulo 367367.

It is a bit more delicate to prove that if we reduce modulo any integer d2d\geq 2, then AA and BB become conjugate under SL2(Z/dZ)\mathrm{SL}_2(\mathbf Z/d\mathbf Z). Stebe's argument runs in two steps.
He first computes the set of matrices VV that conjugate AA to BB, namely he solves the equation VA=BVVA=BV. The answer is given by
 V=V(x,y)=( xy11y25xy). V=V(x,y) = \begin{pmatrix} x & y \\ 11 y & 25 x-y \end{pmatrix}.
Moreover, one has
 det(V(x,y))=25x2xy11y2. \det(V(x,y))=25x^2 - xy -11y^2.
Consequently, to prove that AA and BB are conjugate in SL2(Z/dZ)\mathrm{SL}_2(\mathbf Z/d\mathbf Z), it suffices to find x,yZ/dZx,y\in\mathbf Z/d\mathbf Z such that det(V(x,y))=1(modd)\det(V(x,y))=1 \pmod d.
To prove that they are conjugate by SL2(Z)\mathrm{SL}_2(\mathbf Z), we need to find x,yZx,y\in\mathbf Z such that det(V(x,y))=1\det(V(x,y))=1, and if we agree to be content with a conjugacy by GL2(Z)\mathrm{GL}_2(\mathbf Z), then solutions of det(V(x,y))=1\det(V(x,y))=-1 are also admissible.

Let us first start with the equations modulo dd. By the Chinese remainder theorem, we may assume that d=pmd=p^m is a power of a prime number pp. Now, if p5p\neq 5, we can take y=0y=0 and xx such that 5x=1(modpm)5x=1\pmod {p^m}. If p=5p=5, we take x=0x=0 and we solve yy for 11y2=1(mod5m)-11y^2=1\pmod {5^m}, which is possible since 114(mod5)-11\equiv 4\pmod 5 is a square, and it is easy, by induction (anyway, this is an instance of Hensel's lemma), to produce yy modulo 5m5^m such that y3(mod5)y\equiv 3\pmod 5 and 11y2=1(mod5m)-11y^2=1\pmod{5^m}.

On the other hand, the equation 25x2xy11y2=±125x^2-xy-11y^2=\pm 1 has no solutions in integers.
The case of 1-1 is easy by reduction modulo 33: it becomes x2+2xy+y2=2x^2+2xy+y^2=2, which has no solution since x2+2xy+y2=(x+y)2x^2+2xy+y^2=(x+y)^2 and 22 is not a square modulo 33.
The case of +1+1 is rather more difficult. Stebe treats it by reducing to the Pell equation u2=1101y2+1u^2=1101y^2+1 and shows by analysing the minimal solution to this Pell equation that yy is divisible by 55, which is incompatible with the initial equation.


From a more elaborate point of view, VV is a smooth scheme over Z\mathbf Z that violates the integral Hasse principle. In fact, VV is a torsor under the centralizer of AA, which is a torus, and the obstruction has been studied by Colliot-Thélène and Xu, precisely in this context. However, I did not make the calculations that could use their work to reprove Stebe's theorem.